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3 years ago

Maria thinks the two expressions 2(3a-2) + 4a and 10a - 2 are equivalent. Is she correct? Explain your reasoning.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

yes they are both equal

Step-by-step explanation:

they are both equal because 2(3a-2)+4a simplified is 10a-2 because if you combine like terms you will have this result!

hope i was clear

have a good day! c:

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Choose the ratio that you would use to convert 2.5 centimeters to meters please

WINSTONCH [101]

First of all, 100cm makes 1m. So compare the statements
100cm=1m
2.5cm=Y
I represented the unknown with Y. Cross and multiply. 2.5 will multiply 1 to give 2.5 and Y will multiply 100 to give 100Y. That will be;
100Y=2.5
Divide both sides by 100
Y=2.5/100
But remember; 100cm is 1 meter. So that will be the same as;
Y=2.5cm/1m
which is C
Hope that helped. Have a nice day

6 0

4 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Kitty [74]

Answer: (-2, 5) and (2, -3)

<u>Step-by-step explanation:</u>

Graph the line y = -2x + 1 (which is in y = mx + b format) by plotting the y-intercept (b = 1) on the y-axis and then using the slope (m = -2) to plot the second point by going down 2 and right 1 unit from the first point:

y - intercept = (0, 1) 2nd point = ( -1, 1).

Graph the parabola y = x² - 2x - 3 by first plotting the vertex and then plotting the y-intercept (or some other point):

$y = x^2-2x-3\quad \rightarrow \quad a=1,\ b=-2,\ c=-3\\\\\text{axis of symmetry:}\ x = \dfrac{-b}{2a}\ \longrightarrow \ x=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\\text{y-value of vertex:}\ f(1) = (1)^2-2(1)-3\quad \longrightarrow \quad y = 1 - 2 - 3=-4\\\\\text{y-intercept:}\ f(0)= (0)^2-2(0)-3\ \longrightarrow \ y=0 - 0 - 3 = -3 \\$

vertex = (1, -4) 2nd point (y-intercept) = (0, -3)

<em>see attached</em> - the graphs intersect at two points: (-2, 5) and (2, -3)

8 0

3 years ago

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago

How do you evaluate 3ab

dimaraw [331]

Lets say a=2 and b=5 you now plug in you numbers to get 3x2x5 we know 3x2=6 so now take 6 and multiply it by 5 so 6x5=30 you can also set this equal to 0 3ab=0 then divide by either a or b ill show you both if you divide by a you are left with 3b=a if you divide by b than you get 3a=b

5 0

3 years ago

Ahhhhhhhhhhhhhhhhhhhhhhh

Bess [88]

Answer:

Its 9

Step-by-step explanation:

I did this in 5th grade... All you gotta do is count and there's 4 in the bottom aka 2 rows of 4 and there's 1 block on the top its 9

4+4+1=9

4+4=8

Add a 1 etc.

5 0

3 years ago